CE2-2AC. Buts AE-AC2+CE2; therefore AE2-2AC2. Book II, Again, because EG=GF, EG2=GF2, EG2+GF2=2GF2. But EF2-EG2+GF2; therefore EF2-2GF2-2CD2, because g 47. 1. CD=GF. Since AE2-2AC2, and EF2-2CD2, AE2+EF2 h 34. 1. =2AC2+2CD2. ButƐ AF2=AE2+EF2, and AD2+DF2= AF2, or AD2+DB2-AF2; therefore AD2+DB2-2AC2+ 2CD2. Therefore, if a straight line, &c. Q. E. D.

Otherwise.

"Because AD2=aAC2+CD2+2AC.CD,andDB2+2BC.CD a 4. 2. "=bBC2+CD2=AC2+CD2, by adding equals to equals, b7. 2. “ AD2+DB2+2BC.CD=2AC2+2CD2+2AC.CD; therefore "by taking away the equal rectangles 2BC.CD and 2AC.CD, "there remains AD2+DB2-2AČ2+2CD2. Q. E. D."

PROP. X. THEOR.

IF a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D; the squares of AD, DB are double of the squares of AC, CD.

From the point C drawa CE at right angles to AB, and a 11. 1. make it equal to AC or CB; join AE, EB; through E drawb EF parallel to AB, and through D draw DF parallel b 31 1. to CE. Because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right c 29. 1 angles; and therefore the angles BEF, EFD are less than

Book II.

dCor. 29.1.

e 5. 1.

f 32. 1.

g 15. 1.

c 29. 1.

h 6. 1.

i 34. 1.

k 47. 1.

C

B

two right angles; therefore EB, FD will meetd, if produced
towards B, D. Let them meet in G, and join AG. Because
AC is equal to CE, the angle CEA is equal to the angle
EAC; and the angle
ACE is a right angle;
therefore each of the an-
gles CEA, EAC is half
a right anglef. For the
same reason, each of the
angles CEB, EBC is
half a right angle. Hence
AEB is a right angle.
Because EBC is half a
right angle, DBG is also half a right angle. But BDG is a
a right angle, because it is equal to the alternate angle
DCE; therefore the remaining angle DGB is half a right
angle, and is therefore equal to the angle DBG ; wherefore
also the side DB is equal to the side DG. Again, because
EGF is half a right angle, and the angle at F a right anglei,
the remaining angle FEG is half a right angle, and equal
to the angle EGF; wherefore also the side GF is equal to
the side FE. Now AE2- AC2+CE2-2AC2, becauseEC=
CA; and EG2=* EF2+FG2=2EF2, because EF=FG; and
since EF-CD, EG2=2CD2. Hence AE+EG2=2AC2+
2CD3. Now AG2-AE2+EG2, and AG?=AD2+DG2=AD2
+DB2, because DG=DB; therefore AD2+DB2=2AC2+
2CD2. Wherefore, if a straight line, &c, Q. E. D.

PROP. XI. PROB.

TO divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square of the other part.

Let AB be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part.

Book II.

Upon AB describe the square_ ABDC; bisect AC in E, and join BE; produce CA to F, and make EF equal to EB, and upon AF describe the square FGHA; AB is di- a 46. 1. vided in H, so that the rectangle AB.BH is equal to the b 10. 1. square of AH.

F

Ꮐ

c 3.1.

e 47. 1.

Produce GH to K. Because the straight line AC is bisected in E, and produced to the point F, the rectangle CF.FA together with the square of AE are equal to the square of d 6. 2. EF, or EB, because EF is equal to EB. But the squares of BA, AE are equal to the square of EB; therefore the rectangle CF.FA together with the square of AE are equal to the squares of BA, AE. Take away the square of AE, which is common to both, and the remaining rectangle CF.FA is equal to the square of AB. Now the figure FK is the rectangle CF.FA, for AF is equal to FG; and AD is the square of AB; therefore FK is equal to AD. Take away the common part AK, and the remainder FH is equal to the remainder HD. But HD is the

A

E

K

H

B

D

rectangle AB.BH, for AB is equal to BD; and FH is the square of AH; therefore the rectangle AB.BH is equal to the square of AH. Wherefore the straight line AB is divided in H, so that the rectangle AB.BH is equal to the square of AH. Which was to be done,

Book II.

PROP. XII. THEOR.

a 12. 1.

b 4. 2.

c 47. 1.

IN obtuse angled triangles if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted between the perpendicular and the obtuse angle.

Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawna perpendicular to BC produced. The square of AB is greater than the squares of AC, CB, by twice the rectangle BC.CD.

=

Because the straight line BD is
divided into two parts in the
point C, BD2b BC2+CD2+
2BC.CD; add A D2 to both; then
BD2+AD2=BC2+ CD2+AD2+
2BC.CD. But AB2-BD2+AD3,
and AC2-CD2+AD2; therefore
AB2 BC2+ AC2 + 2BC.CD;
that is, AB2 is greater than BC2+ B
AC by 2BC.CD. Therefore,
in obtuse angled triangles, &c.
Q. E. D.

=

C

A

PROP. XIII. THEOR.

Book II.

IN every triangle the square of the side subtending any of the acute angles is less than the squares of the sides containing that angle, by twice the rect. angle contained by either of these sides and the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle.

Let ABC be any triangle, and the angle at B one of its acute angles; and upon BC, one of the sides containing it, let fall the perpendiculara AD from the opposite angle. The a 12. 1. square of AC, opposite to the angle B, is less than the squares of CB, BA, by twice the rectangle CB.BD.

First, let AD fall within the triangle ABC; then BC2+ BD2=2BC.BD+CD3. Add to each AD; then BC2+BD2+ AD22BC.BD+CD2+ AD3. But BD2+AD-AB, and CD' +DA2="AC2; therefore BC2+ AB2-2BC.BD+AC2; that is, AC2 is less than BC2+AB3 by 2BC.BD.

Secondly, let AD fall without the triangle ABC*; then because the angle at D is a right angle, the angle ACB is greater than a right angle, and AB2=a AC2+BC2+2BC.CD. d 16. 1. Add BC2 to each; then AB+BC2=AÇ2+2BC2+2BC.CD. e 12. 2. But, because BD is divided into two parts in C, BC2+ BC.CD=BC.BD, and 2BC2+2BC.CD=2BC.BD; there- f3. 2. fore AB2+BC-AC2+2BC.BD and AC is less than AB2 +BC by 2BD.BC. Therefore, in every triangle, &c. Q. E. D.

* See the figure of the last proposition.